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Asked: 2024-09-28T00:05:57+05:30 In:

How can I build a concise integer operation calculator in Python without using eval()?

I’ve been diving into some interesting challenges lately, and I stumbled upon this cool problem related to building a simple integer operation calculator that can handle basic arithmetic in a unique way. Here’s the deal: the objective is to create a program that takes a math expression as a string input and evaluates it using integer operations, but you have to do it in the most concise way possible. Think of it as a sort of puzzle where brevity is key!

Here’s where it gets a bit tricky: the operations you’re allowed to use are addition, subtraction, multiplication, and division. Oh, and it gets more interesting! You have to deal with potential spaces between numbers and operators, which means your program needs to be smart enough to parse the input correctly before doing any calculations. Also, you can’t rely on any built-in functions that directly evaluate strings like `eval()`, so it’s all about the clever manipulation of the input.

To make things even more challenging, consider how you can optimize your solution in terms of character count. I mean, surely there’s a way to accomplish the task with fewer characters than others, right? I’ve seen some pretty impressive solutions that use clever algorithms or data structures to minimize the size of their code while still producing the correct output.

So here’s my question to you all: how would you approach this challenge? What techniques or tricks would you use to create your integer operation calculator? And can you think of any edge cases that might trip up a simpler implementation?

I’m really curious to hear your thoughts and strategies on tackling this problem. If you’ve already had a go at it, feel free to share your solution or even some snippets that you think are clever! Let’s brainstorm some code golf ideas together. Can’t wait to see what you come up with!

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    2 Answers

    1. To tackle this integer operation calculator challenge, I would first create a function that processes the string input effectively by removing any unnecessary spaces and then parsing the expression into tokens of numbers and operators. Since we can’t use built-in evaluation functions like `eval()`, I would implement the Shunting Yard algorithm, which can convert the input expression into Reverse Polish Notation (RPN). This step allows for the orderly evaluation of the expression using a stack-based approach, handling the precedence of operations correctly. The following concise code illustrates this idea:

      def calc(s): 
 s=s.replace(' ',''); 
 def pop(): 
  return stack.pop(); 
 stack, postfix, o={}(1),[] 
 for c in s+('$'): 
  if c.isdigit(): 
   n=''; 
   while c.isdigit(): 
    n+=c; 
    c=s[s.index(c)+1] 
   postfix.append(int(n)) 
  elif c in"+-": 
   while len(stack) > 1: 
    postfix.append(pop()) 
   stack[0]=c 
  elif c in '*/': 
   while len(stack) > 1 and stack[-1] in '*/': 
    postfix.append(pop()) 
   stack.append(c) 
 for o in postfix: 
  if o in"+-*/": 
   b,a=pop(),pop(); 
   stack.append(a+b if o=="+" else a-b if o=="-" else a*b if o=="*" else a//b) 
 return stack[0]

      In this implementation, the function `calc` uses two main components: a stack to keep track of operators and a list to form the postfix expression. After constructing the postfix notation, I evaluate it using a stack to ensure that operations are performed in the correct order. One edge case to consider is the division by zero, which is handled by checking the divisor before performing the division operation. This approach combines efficiency and brevity, fulfilling the puzzle’s requirements effectively.

    2. def calc(s):
 s = s.replace(' ','')
 num,stack,op = '',[],'+'
 for i in range(len(s)+1):
  if i==len(s) or s[i] in '+-*/':
   if num: stack.append(int(num))
   if op=='-': stack.append(-stack.pop())
   if op=='*': stack[-1]*=int(num)
   if op=='/': stack[-1]//=int(num)
   op,num = s[i],'' 
 return sum(stack)

# Example usage
print(calc("3+5 / 2 "))  # Output: 5

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